We have,

By applying C₁→ C₁ + C₂ + C₃, we get

Taking (12 + x) common from the first column, we get

By applying C₂→ C₂ + C₃, we get

Applying R₂→ R₂ – R₃, we get

Applying R₃→ R₃ – R₁, we get

Expanding along first column, we get

⇒ (12 + x)[(1){0 – (2x)(-2x)}] = 0

⇒ (12 + x)(4x²) = 0

⇒ 12 + x = 0 or 4x² = 0

⇒ x = -12 or x = 0

Hence, the value of x = -12 and 0